Chemistry Homework Answer Key #9
Chapter 12 - Electrochemistry
19. The galvanic cell given is: Ni(s) | Ni2+ (aq) || Ag+ (aq) | Ag (s) E°= 1.03 V
oxidation: Ni (s) ® Ni2+ (aq) + 2 e-reduction: Ag+ (aq) + e-® Ag (s)
Net reaction: Ni (s) + 2 Ag+ (aq) ® Ni2+ (aq) + 2 Ag (s)
In this redox reaction, n = 2.
23. In this problem, solid indium is observed to plate out on one
of the electrodes as the galvanic cell operates. Since this is a reduction
product, this electrode must be the cathode and the reduction half-reaction
is:
reduction: In3+ (aq) + 3 e-® In (s).The oxidation is therefore the generation of Zn2+ from zinc at the anode, or,
oxidation: Zn (s) ® Zn2+ (aq) + 2e- (aq)
Net reaction: 3 Zn (s) + 2 In3+ (aq) ® 3 Zn2+ (aq) + 2 In (s)
The observed cell potential (all species under standard conditions) is E°= 0.425 V.
Since DE°= E°red + E°ox, we can find E°red by the following (Eox is the same in magnitude but opposite in sign of the standard reduction potential of Zn2+, or + 0.763 V):
E°red = DE°- E°ox
E°red = 0.425 V - (0.763 V) = -0.338 V.
So, we could write: In3+ (aq) + 3 e- ® In (s) E°= -0.338 V
35. The half-reactions is this cell are:
oxidation: Cr2+ (aq) ® Cr3+ (aq) + e- Eox = -(-0.424 V)reduction: Pb2+ (aq) + 2 e-® Pb (s) Ered = -0.126 V
Net reaction: 2 Cr2+ (aq) + Pb2+ (aq) ® Pb (s) + 2 Cr3+ (aq)
DE°= E°red + E°ox = 0.424 - 0.126 V = 0.298 V
n = 2
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The actual potential of the cell is given by the Nernst Equation:
39. In this application of the Nernst Equation, you must first determine
the value of Q, and from there calculate the concentration of the unknown
species. The half-reactions are:
oxidation: H2 (g) ® 2 H+ (aq) + 2e- E°ox = 0.000 Vreduction: I2 (s) + 2 e-® 2 I- (aq) E°red = 0.535 V
net reaction: H2 (g) + I2 (s) ® 2 H+ (aq) + 2 I- (aq) DE°= 0.535 V
n = 2
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log Q = -10.35; Q = 4.48 ´ 10-11
=
[H+] = 6.69 ´ 10-6; pH = 5.17
41. The half-reactions and corresponding standard potentials for
this reaction are:
oxidation: 7 H2O (l) + 2 Cr3+ (aq) ® Cr2O72- (aq) + 14 H+ (aq) + 6 e- E°ox = - (1.33 V)reduction: 2 e- + 2 H+ (aq) + HClO2 (aq) ® HClO (aq) + H2O (l) E°red = 1.64 V
DE°= E°red + E°ox = 1.64 - 1.33 V = 0.31 V
In the overall balanced redox reaction, n = 6.
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log Q = 16.23; Q = 1.71 ´ 1016
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[Cr3+] = 1.05 ´ 10-8
47. This is an example of a concentration cell, that is, a cell
in which a potential exists only because of a difference in concentration
of the same species at the two electrodes. The half-reactions are:
oxidation: H2 (g) ® 2 H+ (aq) + 2e- E°ox = 0.000 Vreduction: 2 H+ (aq) + 2e-® H2 (g) E°red = 0.000 V
DE° = 0.000
There is no net reaction in this cell. There is however a reduction of H+ taking place at the cathode and an oxidation of H2 at the anode. This can be written as:
2 H+ (aq, cathode) + H2 (g, anode) ® H2 (g, cathode) + 2 H+ (aq, anode)
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Using the Nernst equation,
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log Q = -5.07; Q = 8.46 ´ 10-6
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[H+ anode] = 2.91 ´ 10-3 M; pH = 2.54
Since this hydronium ion concentration is maintained by a buffer in which [HA] = 0.10 M and [A-] = 0.10 M, the Henderson-Hasselbach equation can be used to find pKa.
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pKa = 2.54; Ka = 2.91 ´ 10-3
49. The half-reactions are:
oxidation: H2 (g) ® 2 H+ (aq) + 2e- E°ox = 0.000Vreduction: Br2 (l) + 2e-® 2 Br- (aq) E°red = 1.065V
net reaction: H2 (g) + Br2 (l) ® 2 Br- (aq) + 2 H+ (aq) DE°= 1.065 V
n = 2
Using the Nernst Equation:
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log Q = -21.81; Q = 1.54 ´ 10-22
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[Br-] = 1.24 ´ 10-11 M
Ksp = [Ag+] [Br-] = (0.060) (1.24 ´ 10-11) = 7.4 ´ 10-13
68. Since silver plates out in this cell as cobalt dissolves, we
can conclude that silver serves as the cathode and cobalt serves as the
anode. The cell notation is then,
Co (s) | Co2+ (aq) || Ag+ (aq) | Ag (s)
71. The reaction at the cathode of this electrolytic cell is: Zn2+ (aq) + 2 e- ® Zn (s)
The half-reactions are:oxidation: Co (s) ® Co2+ (aq) + 2 e-
reduction: Ag+ (aq) + e-® Ag (s)
net reaction: Co (s) + 2 Ag+ (aq) ® Co2+ (aq) + 2 Ag (s)
b) This is a disguised simple stoichiometry problem:
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c) The average current that flowed through the cell is given by:
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73. The reduction of Al2O3 to Al can be written
as:
Al2O3 (s) + 6 e-® 2 Al (s) + 3 O2-75. The half-reactions are:DG = -nFE
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Ouch!
MnO2 (s) + 4 H+ (aq) + 2e- ® Mn2+ (aq) + 2 H2O (l) E°red = +1.208 VO2 (g) + 2 H+ (aq) + 2e-® H2O2 (aq) E°red = +0.682 V
Since one of these reactions must run backwards, that is, as an oxidation, we must determine which of these two half-reaction we should reverse. The cell is a galvanic cell, so DE > 0.
Since DE° = E°ox + E°red, the oxidation reaction (at the anode) must be:
H2O2 (aq) ® O2 (g) + 2 H+ (aq) + 2e-E°ox = -0.682 V
The cathode reaction (the reduction) is therefore:
MnO2 (s) + 4 H+ (aq) + 2e-® Mn2+ (aq) + 2 H2O (l)
b) DE° = -0.682 + 1.208 = 0.526 V